Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 2 - Section 2.6 - Limits at Infinity; Horizontal Asymptotes - 2.6 Exercises: 18

Answer

$\lim\limits_{x\to-\infty}\frac{4x^3+6x^2-2}{2x^3-4x+5}=2$

Work Step by Step

$$A=\lim\limits_{x\to-\infty}\frac{4x^3+6x^2-2}{2x^3-4x+5}$$ Divide both numerator and denominator by $x^3$ $A=\lim\limits_{x\to-\infty}\frac{4+\frac{6}{x}-\frac{2}{x^3}}{2-\frac{4}{x^2}+\frac{5}{x^3}}$ $A=\frac{\lim\limits_{x\to-\infty}(4+\frac{6}{x}-\frac{2}{x^3})}{\lim\limits_{x\to-\infty}(2-\frac{4}{x^2}+\frac{5}{x^3})}$ $A=\frac{\lim\limits_{x\to-\infty}4+6\lim\limits_{x\to-\infty}\frac{1}{x}-2\lim\limits_{x\to-\infty}\frac{1}{x^3}}{\lim\limits_{x\to-\infty}2-4\lim\limits_{x\to-\infty}\frac{1}{x^2}+5\lim\limits_{x\to-\infty}\frac{1}{x^3}}$ $A=\frac{4+6\times0-2\times0}{2-4\times0+5\times0}$ $A=2$
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